A C voltage V ( t ) = 20   sin ω t of frequency 50   Hz is applied to a parallel plate…

From the given information,

$C=\frac{{\in }_0 A}{ d}=\frac{{ϵ}_0\times 1}{2\times {10}^{-3}} F$
$\therefore X_{\mathrm{C}}=\frac{1}{\omega C}=\frac{2\times {10}^{-3}}{2\times 50\pi \times {\in }_0}=\frac{2\times {10}^{-3}}{25\times 4\pi {\in }_0} \mathrm{\Omega }$
$\therefore X_{\mathrm{C}}=\frac{2\times {10}^{-3}}{25}\times 9\times {10}^9=\frac{18}{25}\times {10}^6 \mathrm{\Omega }$
$\therefore i_0=\frac{V_0}{X_{\mathrm{C}}}=\frac{20\times 25}{18}\times {10}^{-6} \mathrm{A}=27.47 \text{μA}.$

The value of amplitude of displacement current will be same as value of amplitude of conventional current.
Hence option $3.$